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<p><dfn class="terminology">Solution:</dfn></p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
(x-1) \frac{Q(x)}{P(x)}=(x-1) \frac{-2 x}{1-x^2}=-2 \frac{x}{1+x},
\end{equation*}
</div>
<p class="continuation">which are analytic at <span class="process-math">\(x=1\)</span> because the rational function has no singularities at <span class="process-math">\(x=1\text{.}\)</span> Besides,</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
(x-1)^2 \frac{R(x)}{P(x)}=(x-1)^2 \frac{\alpha (\alpha+1)}{1-x^2}=(1-x) \frac{\alpha (\alpha+1)}{1+x},
\end{equation*}
</div>
<p class="continuation">which is also analytic at <span class="process-math">\(x=1\text{.}\)</span> In conclusion, <span class="process-math">\(x=1\)</span> is a regular singular point.Note: In this course, we will not consider the case of irregular singular points.</p>
<span class="incontext"><a href="sec5_2.html#p-212" class="internal">in-context</a></span>
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